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In my research on linear algebra and its applications, I have come across the following problem which has stumped me:

Let $ A $ be a positive definite matrix and let $ D $ be a positive diagonal matrix with entries on the main diagonal: $ d_1,d_2,...,d_n $, both $ A $ and $ D $ have the same dimension $ n \times n $. I was interested in understanding how the eigenvalues of the sum $ A + ADA $ qualitatively behave with respect to the eigenvalues of $ A $ and the entries $ d_1,...,d_n $.

I thought since this sum has a special form, one could hopefully say a bit more analytically than by exclusively using the fact that the two summands are Hermitian. I thought about using some other techniques, perhaps expressing the product of $A$ and $ D $ as a polynomial of $ A $, but unfortunately, I am stuck. I certainly appreciate all help on this.

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    $\begingroup$ Because $A>0$ on a finite dimensional space we have no problems to build the (bounded) operator / matrix $A^{-1}>0$. Then $$A+ADA=A(A^{-1}+D)A\ ,$$ so the problem can be restated: Given two diagonalizable matrices $B,D>0$, (and without restriction, let $D$ be already diagonal,) what can be said about the spectrum of the sum in terms of the spectra of $B,D$? Answer: Well, not so much. Does this answer the question? $\endgroup$
    – dan_fulea
    Commented Apr 30, 2018 at 20:20
  • $\begingroup$ @dan_fulea yes that answers it thanks, I thought it might be too intracate to tackle $\endgroup$
    – groupoid
    Commented Apr 30, 2018 at 20:35
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    $\begingroup$ @dan_fulea I don't follow your argument --- how is $B$ related to $A$ and $D$? $\endgroup$
    – Federico Poloni
    Commented Apr 30, 2018 at 22:10
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    $\begingroup$ @DimaPasechnik But then the eigenvalues of $B+D$ aren't related in a simple way to those of $A(B+D)A$... $\endgroup$
    – Federico Poloni
    Commented May 1, 2018 at 20:02
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    $\begingroup$ Yes, $B$ was supposed to be the inverse of $A$. Let then $B+D$ have an eigenvector $w$ for some eigenvalue $\lambda$, so $(B+D)w=\lambda w$. We write $w=Av$ for the one $v$. Then $(B+D)Av = \lambda Av$. And there is no "algebraic" way to relate $\lambda$ to an eigenvalue of $A(B+D)A$, sorry. Thanks for the comment, Federico! $\endgroup$
    – dan_fulea
    Commented May 2, 2018 at 10:55

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